Question

Use Implicit Differentiation to determine the equation of a line tangent to x + xy = 2y at the point (3, 2).

Answer 1

Given:

`x+xy=2y`

You need to use Implicit Differentiation to determine the equation of a line tangent. In order to do this, you need follow these steps:

1. Derivate the function treating "y" as function of "x". Use these Derivative Rules:

- Power Rule:

`(d)/(dx)(nx^n)=nx^(n-1)`

- Product Rule:

`(d)/(dx)(u(x)\cdot v(x))=u^(\prime)(x)^v(x)+v^(\prime)(x)u(x)`

Then, you get:

`(x)^(\prime)+(xy)^(\prime)=(2y)^(\prime)`
`(1x^(1-1))^(\prime)+(x)^(\prime)(y)+(y)^(\prime)(x)=2y^(\prime)`
`1+(1)(y)+xy^(\prime)=2y^(\prime)`
`1+y+xy^(\prime)=2y^(\prime)`

2. Solve for y':

`1+y=2y^(\prime)-xy^(\prime)`
`1+y=y^(\prime)(2-x)`
`y^(\prime)=(1+y)/(2-x)`
`y^(\prime)=-(1+y)/(x-2)`

3. Having the point:

`(3,2)`

You need to substitute its coordinate into the derivative of the function and evaluate, in order to find the slope of the line that is tangent to the given function:

`y^(\prime)=-(1+(2))/((3)-2)=-(3)/(1)=-3`

4. The Point-Slope Form of the equation of a line is:

`y-y_1=m(x-x_1)`

Where "m" is the slope and this is a point on the line:

`(x_1,y_1)`

Substituting values, you get:

`y-2=-3(x-3)`

5. You can rewrite it in Slope-Intercept Form by solving for "y":

`y=-3x+9+2`
`y=-3x+11`

Hence, the answer is:

`y=-3x+11`